The recent question Launch windows to Mars got me thinking. Why is there only one or two launch windows to Mars ever other year?

If you leave Earth heading for the sun, you should get a gravity assist on flyby.

At least in my mind, the travel time to Mars would be the same no matter where the two plants are. I imagine that the fastest anything man made ever went was on a powered trajectory towards Sol. The same amount of fuel needed to get out to Mars, must be enough to get to the Mars via the Sun.

  • 2
    $\begingroup$ One actually has to put in a great delta-v effort in getting close to the Sun to begin with. Krafft Arnold von Ericke (one of Wernher von Braun's old friends) suggested sending a spacecraft way out to Jupiter to halt its orbital speed (by gravity "resist") so that it could fall down towards the Sun. Gravity assist is only available as a velocity direction changer. As in the diagonal being a bit longer than the side of a rectangle, to put it in a way I can understand it. Since we are basically co-moving with the Sun, there's not much gravity assist to be gained there in interstellar terms. $\endgroup$
    – LocalFluff
    Nov 4, 2016 at 18:36
  • 1
    $\begingroup$ Besides the immense delta-v needed to get on a close solar approach, the relative speed between Mars and the spacecraft would be vastly higher at arrival; you'd be approaching essentially perpendicular to Mars's orbital speed around the sun, about 24km/s on average. Fine for a very fast flyby, but prohibitive if you want to orbit Mars or land in one piece. $\endgroup$ Nov 4, 2016 at 18:49
  • $\begingroup$ The fastest anything man made ever went to Pluto. en.wikipedia.org/wiki/New_Horizons $\endgroup$ Nov 4, 2016 at 19:30
  • $\begingroup$ Obligatory xkcd $\endgroup$
    – Cody
    Nov 4, 2016 at 20:08
  • $\begingroup$ A solar flyby is not possible because of the radiation of the sun. The very intense radiation of a close flyby would destroy the space ship. If the distant to sun is reduced to 1/10 of the distance sun to earth, the radiation power rises by a factor of 100. $\endgroup$
    – Uwe
    Nov 13, 2016 at 12:24

1 Answer 1


If you leave Earth heading for the sun, you should get a gravity assist on flyby.

When you travel in space, strictly speaking all masses attract each other (n-body-problem). However most time, there is one central body, whose gravity is dominant and exceeds other forces by several orders of magnitude. For traveling between planets in our solar system, this is clearly the sun.

A gravity-assist or slingshot means you approach another body, so its gravity exceeds the suns for a short time, the dominant central body changes. If you look at this in a sun reference frame, this bends your trajectory. The trick of this maneuver is, that in the assisting planets frame, you do not loose or gain energy, but in the suns frame you do, because the frames a moving against each other.

Can our sun itself assist you? No, it can't, because you are already under it's influence, the trick with changing the dominant body does not work here (unless you come from outside our solar system).

It happens that also other assists (except luna) will not help, because reaching the assisting body needs more energy than reaching mars directly. A Hohmann-transfer is the cheapest way to get there.

At least in my mind, the travel time to Mars would be the same no matter where the two plants are.

Now Hohmann will arrange for getting to mars' orbit, but theres nothing to be found there unless mars happens to be right at this point when you arrive. Braking there and waiting until mars arrives does also not work, you need to keep your speed relative to the sun in order to stay there. So you and mars would be traveling at the same speed and never meet. Starting at the right time is essential.

The same amount of fuel needed to get out to Mars, must be enough to get to the Mars via the Sun.

No, Walter Hohmann showed in his work The attainability of heavenly bodies that this is not the case. In order to approach sun, you need to slow down in suns reference frame to lower centrifugal forces. To get to mars, you need to accelerate, because it is farther out.

Is the fastest way...?

If you can afford the thrust all the time, a direct intercept to your target is always fastest. The pointing of your thrusters needs to be tilted somewhat to counteract the Coriolis force. Low consumption thrusters strong enough to escape from earth are not available (as of 2016).

With limited propellants, mission duration will be shortest if you use them in the beginning and at the end, which again means a Hohmann trajectory.

Earliest arrival after a given point in time is different, because you will not be able to delay your start until a good launch opportunity. For earliest arrival, you may need a completely different trajectory. Gravity-assists may be beneficial in this case to leave the ecliptic.

  • $\begingroup$ What you can do though is fire up your engines on closest approach to the Sun to get way more bang for your buck in increasing your orbital energy. Assuming you don't get fried in the process. $\endgroup$
    – Mark Adler
    Nov 5, 2016 at 1:20
  • $\begingroup$ s/breaking/braking/ $\endgroup$
    – hobbs
    Nov 5, 2016 at 1:47
  • $\begingroup$ @MarkAdler Isn't the point of this answer that there's no point in having your "closest approach to the Sun" be any time other than when you leave Earth? $\endgroup$
    – hobbs
    Nov 5, 2016 at 1:49
  • $\begingroup$ Hohmann is the cheapest trajectory, but not the fastest. OP is asking about the fastest. $\endgroup$ Nov 5, 2016 at 1:59
  • 1
    $\begingroup$ @Andreas my understanding is that that's the kind of thing that takes serious computation. The case where you can treat your burns as impulses is child's play by comparison :) $\endgroup$
    – hobbs
    Nov 5, 2016 at 2:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.